Re: Problem with change of variables in an integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg131574] Re: Problem with change of variables in an integral*From*: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>*Date*: Thu, 5 Sep 2013 08:13:38 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net

Hello, Although I know how to make a change of variables in an integral I can only do it manually by applying a substitution rule to the integrand and the differential e.g {f[z],\[DifferentialD]z}//. {z-> r E^(I \[Phi]),\[DifferentialD]z->E^(I \[Phi]) \[DifferentialD]r,\[Phi] -> (2\[Pi]/3} But it cannot applied this substitution rule directly to the integral, e.g. Integrate[f[z],{z,0,\[Infinity]}] //. {z-> r E^(I \[Phi]),\[DifferentialD]z->E^(I \[Phi]) \[DifferentialD]r,\[Phi] -> (2\[Pi]/3} Comparing with the correct result, the exponential factor E^((2 I \[Pi])/3) = (-1)^(2/3) is missing in the evaluation of the integral. The correct appearance of the integral is : Integrate[1/(1+r^3) E^((2 I \[Pi])/3),{r,0,\[Infinity]}] How can I force Mathematica (V8) to perform the correct transformation of variables as regards to the integral (and not to its separate parts of it as {f[z],\[DifferentialD]z} ? Any suggestions are appreciated. Robert Kragler Hello, Robert, Try this: Clear[f, \[Phi]]; Map[ReplaceAll[#, {z -> r*Exp[I \[Phi]] , \[Phi] -> 2 \[Pi]/3}] &, Integrate[f[z], {z, 0, \[Infinity]}]] to map the replacements directly onto the structures below the Integrate operator. The result is below: \!\( \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(f[ \*SuperscriptBox[\(E\), \(I\ \[Phi]\)]\ r] \[DifferentialD]\(( \*SuperscriptBox[\(E\), \(I\ \[Phi]\)]\ r)\)\)\) If it is used, for example, with the function f[z_]:=1/(1+z^3), as you tried: Clear[f, \[Phi]]; f[z_] := 1/(1 + z^3); Map[ReplaceAll[#, {z -> r*Exp[I \[Phi]] , \[Phi] -> 2 \[Pi]/3}] &, Integrate[f[z], {z, 0, \[Infinity]}]] it gives (2 \[Pi])/(3 Sqrt[3]) When I do the same "by hand": \[Phi] = 2 \[Pi]/3; Integrate[Exp[I \[Phi]]/( 1 + (Exp[I \[Phi]]*r)^3), {r, 0, \[Infinity]}] I get this: (2 (-1)^(2/3) \[Pi])/(3 Sqrt[3]) which in my opinion is almost the same, though not entirely. I do not see, where from the difference comes. Surprisingly strange result has been delivered, when I used Esc+e+e+Esc instead of Exp to check of how the "hand-made" transformation would work: \[Phi] = 2 \[Pi]/3; Integrate[E^(i \[Phi])/(1 + (E^(i \[Phi]) r)^3), {r, 0, \[Infinity]}] The answer was: ConditionalExpression[(2 E^((2 i \[Pi])/3) \[Pi])/( 3 Sqrt[3] (E^(2 i \[Pi]))^(1/3)), Im[E^(-2 i \[Pi])] != 0 || Re[E^(-2 i \[Pi])] >= 0] which I do not understand. I do it on PC, Win XP, Mathematica 9.0.1.0. I would be v= ery interested in comments of the Community on this matter. Best, Alexei Alexei BOULBITCH, Dr., habil. IEE S.A. ZAE Weiergewan, 11, rue Edmond Reuter, L-5326 Contern, LUXEMBOURG Office phone : +352-2454-2566 Office fax: +352-2454-3566 mobile phone: +49 151 52 40 66 44 e-mail: alexei.boulbitch at iee.lu